Compute the stochastic differential for $t\cdot B_t$ using Ito's Lemma

Question

Consider $Y_t=B_t^3$, $t \geq 0$ where $(B_t)_{t \geq 0}$ is standard Brownian Motion.

Here, $dB_t = 0+1\cdot dB_t$ and $G(x,t)=x^3$, What is $dY_t$?

Using Ito's Lemma, we can calculate the partial derivatives and get:

$$\begin{align*} dY_t &= [3B_t^3\cdot 0+0\cdot \frac{1}{2}\cdot 6B_t\cdot 1^2]dt + 3B_t^2\cdot 1\cdot dB_t \\ &= 3B_t^2dt + 3B_t^2dB_t \end{align*}$$

If $Y_t = t\cdot B_t$, how can I use this same logic to get $d(tB_t)$?

Answer 1

You need to use Itô's lemma with $g(x,t) = xt$. Notice that $g_x(x,t) = t, g_t(x,t) = x, g_{xx}(x,t) = 0$, so that \begin{align*} dY_t &= dg(B_t, t) \\ &= g_t(B_t, t) dt + g_x(B_t, t) + \frac{1}{2}g_{xx}(B_t,t)(dB_t)^2 \\ &= B_t dt +t dB_t \end{align*}